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Lemma. Suppose $G$ is an indecomposable group which satisfies both chain conditions. Let $f$ and $g$ be normal nilpotent endomorphisms of $G$ and suppose that $f+g$ is also an endomorphism. Then $f + g$ is nilpotent.

Recall that an endomorphism $f$ is called nilpotent if there exists a positive integer $k$, such that $f^k = \textbf{0}$, where $\textbf{0}$ denotes the endomorphsim, which sends all elements of $G$ to the identity.

Please give some valuable hints

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  • $\begingroup$ You use the notation $f+g$, so is $G$ abelian? By identity, do you then mean $0 \in G$? Judging from the title this has to do something with Krull Schmidt; could you elaborate on this? $\endgroup$
    – Jendrik Stelzner
    Commented Jan 5, 2016 at 10:12
  • $\begingroup$ How can i edit my question ? I have to also add that f and g are also nilpotent . Its a lemma to the krull schmidt theorem . G need no to be abelian i have mentioned that if f+g is endomorphism then we are interseted in this lemma $\endgroup$
    – Mathslover shah
    Commented Jan 5, 2016 at 10:28
  • $\begingroup$ Under your question there should be a link saying "edit". My problem is that if $G$ is not abelian then I don’t understand how the endomorphism $f+g$ is defined; for abelian groups I understand it as $(f+g)(x) = f(x)+g(x)$. $\endgroup$
    – Jendrik Stelzner
    Commented Jan 5, 2016 at 10:30
  • $\begingroup$ It is defined as f+g(x) = f(x)g(x) $\endgroup$
    – Mathslover shah
    Commented Jan 5, 2016 at 10:34
  • $\begingroup$ The book uses its prevoious result which says that if G is indecomposable having both chain conditions then every normal endomorphism is either nilpotent or automorphism so it supposes that let f+g is an automorphism led to a contradiction hence we get the other possibility which is the proof of this lemma but i am stucked there so may be someone approaches deferently $\endgroup$
    – Mathslover shah
    Commented Jan 5, 2016 at 10:41

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